Question

find the limit.
lim(x→0) (5 + 4x+sin x)/(6 cos x)

simplify the expression inside the given limit. select the correct choice below and, if ne
a.
lim(x→0) (5 + 4x+sin x)/(6 cos x)=lim(x→0) ( ) (simplify your answer.)
b. the expression inside the limit cannot be simplified.

Explanation:

Step1: Use limit - sum and quotient rules

We know that $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$ (where $\lim_{x
ightarrow a}g(x)
eq0$) and $\lim_{x
ightarrow a}(f(x)+g(x)+h(x))=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)+\lim_{x
ightarrow a}h(x)$. So, $\lim_{x
ightarrow0}\frac{5 + 4x+\sin x}{6\cos x}=\frac{\lim_{x
ightarrow0}(5 + 4x+\sin x)}{\lim_{x
ightarrow0}(6\cos x)}=\frac{\lim_{x
ightarrow0}5+\lim_{x
ightarrow0}(4x)+\lim_{x
ightarrow0}\sin x}{6\lim_{x
ightarrow0}\cos x}$.

Step2: Evaluate individual limits

We know that $\lim_{x
ightarrow0}c = c$ (where $c$ is a constant), $\lim_{x
ightarrow0}x = 0$, $\lim_{x
ightarrow0}\sin x=0$ and $\lim_{x
ightarrow0}\cos x = 1$. So, $\frac{\lim_{x
ightarrow0}5+\lim_{x
ightarrow0}(4x)+\lim_{x
ightarrow0}\sin x}{6\lim_{x
ightarrow0}\cos x}=\frac{5 + 4\times0+0}{6\times1}$.

Step3: Simplify the result

$\frac{5 + 0+0}{6}=\frac{5}{6}$.

For the simplification part:
The expression inside the limit $\frac{5 + 4x+\sin x}{6\cos x}$ cannot be further simplified in a non - trivial algebraic way before applying the limit rules. So the answer for the simplification choice is B. The expression inside the limit cannot be simplified.

Answer:

The value of $\lim_{x
ightarrow0}\frac{5 + 4x+\sin x}{6\cos x}$ is $\frac{5}{6}$; For the simplification choice: B. The expression inside the limit cannot be simplified.